A CST element has nodal coordinates (10,10).(70,35) and (75,25) for nodes 1.2 and 3 respectively The element is 2 mm thick and is of material with properties E =70 GPA. poission's ratio is 0.3 After applying the load to the element the nodal deformation were found to be u1 = 0.01 mm. v1 = -0.04 mm. u2 = 0.03, v2 =0.02 mm. u3 = -0.02 v3= -0.04 mm. Determine the strains ex , ey , exy and corresponding .

$\beta_1=y_2-y_3 = 10 \hspace{2cm}r_1 = x_3-x_2 = 5$

$\beta_2=y_3-y_1 = 15 \hspace{2cm}r_2 = x_1-x_3 = 65$

$\beta_3=y_1-y_2 = 25 \hspace{2cm}r_3 = x_2-x_1 = 60$

[B] = $\frac{1}{2A} \begin{bmatrix} \ \beta_1 & 0 & \beta_2 & 0 & \beta_3 & 0 \\ \ 0 & r_1 & 0 & r_2 & 0 & r_3 \\ \ r_1 & \beta_1 & r_2 & \beta_2 & r_3 & \beta_3 \\ \end{bmatrix} $

${2A} \begin{vmatrix} \ \ 1 & x_1 & y_1 \\ \ 1 & x_2 & y_2 \\ \ 1 & x_3 & y_3 \\ \end{vmatrix} = \begin{vmatrix} \ \ 1 & 10 & 10 \\ \ 1 & 70 & 35 \\ \ 1 & 75 & 25 \\ \end{vmatrix}=-725$

[e] = {B} {u}

$\begin{Bmatrix} \ e_x \\ \ e_y \\ \ r_{xy} \\ \end{Bmatrix} = \frac{1}{-725}\begin{bmatrix} \ 10 & 0 & 15 & 0 & -25 & 0 \\ \ 0 & 5 & 0 & -65 & 0 & 60 \\ \ 5 & 10 & -65 & 15 &6 0 & -25 \\ \end{bmatrix}\begin{Bmatrix} \ 0.01 \\ \ -0.04 \\ \ 0.03 \\ \ 0.02 \\ \ -0.02 \\ \ -0.04 \\ \end{Bmatrix}$

$e_x = -0.1448 \times 10^{-2}$

$e_y = -0.5378 \times 10^{-2}$

$e_{xy} = 0.3035 \times 10^{-2}$

Elemental stress,

$[\sigma]$ = [D] {e}

[D] = $\frac{E}{1-v^2}\begin{bmatrix} \ 1 & v & 0 \\ \ v & 1 & 0 \\ \ 0 & 0 & \frac{1-v}{2} \\ \end{bmatrix} = \frac{0.7 \times10^5}{1-0.09} \begin{bmatrix} \ 1 & 0.3 & 0 \\ \ 0.3 & 1 & 0 \\ \ 0 & 0 & 0.35 \\ \end{bmatrix}$

= $10^3\begin{bmatrix} \ 76.92 & 23.077 & 0 \\ \ 23.077 & 76.92 & 0 \\ \ 0 & 0 & 26.92 \\ \end{bmatrix}$

$ \therefore \begin{Bmatrix} \ \sigma_x \\ \ \sigma_y \\ \ z_{xy} \\ \end{Bmatrix} = \begin{bmatrix} \ 76.92 & 23.077 & 0 \\ \ 23.077 & 76.92 & 0 \\ \ 0 & 0 & 26.92 \\ \end{bmatrix}\begin{Bmatrix} \ -1.448 \\ \ -5.378 \\ \ 3.035 \\ \end{Bmatrix}$

$\sigma_x=235.54 N/mm^2$

$\sigma_y = -447.72 $

$z_{xy} = 81.87 N/mm^2$